University of Florida/Egm4313/s12.team8.dupre/R6.1

R6.1

Problem Statement

Given $\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos(n\omega x)+b_{n}sin(n\omega x)]$:

(1) Find the (smallest) period of $\displaystyle cos(n\omega x)$ and $\displaystyle sin(n\omega x)$.

(2) Show that these functions also have period $\displaystyle p$.

(3) Show that the constant $\displaystyle a_{0}$ is also a periodic function with period $\displaystyle p$.

Solution (1)

We know that the period of a normal $\displaystyle sin(x)$ or $\displaystyle cos(x)$ is $\displaystyle 2\pi$. When there are values or variables being multiplied by this $\displaystyle x$ variable, the period becomes $\displaystyle 2\pi$ divided by the values or variables. We know that:

$\displaystyle \omega ={\frac {\pi }{L}}={\frac {2\pi }{p}}$ (6.1.1)

Where $\displaystyle p$ is the period.

Using this relation, along with our variable $\displaystyle n$, we can solve for the period of $\displaystyle cos(n\omega x)$ and $\displaystyle sin(n\omega x)$ as follows:

$\displaystyle p={\frac {2\pi }{\tfrac {n\pi }{L}}}$

$\displaystyle p={\frac {L\pi }{n}}$ (6.1.2)

Since the period will be smallest at $\displaystyle n=1$, plugging into equation (6.1.2) shows that the smallest period of $\displaystyle cos(n\omega x)$ and $\displaystyle sin(n\omega x)$ is:

$\displaystyle p=L\pi$     (6.1.3)

Solution (2)

We are also given that:

$\displaystyle \omega ={\frac {2\pi }{p}}$ (6.1.4)

Using this relation, we can solve for the period again as follows:

$\displaystyle p={\frac {2\pi }{\frac {n2\pi }{p}}}$

$\displaystyle p={\frac {p}{n}}$ (6.1.5)

We know that the period is smallest at $\displaystyle n=1$ , and plugging this value into (6.1.5) proves that:

$\displaystyle p=p$     (6.1.6) 

Solution (3)

We know that, starting at 0:

$\displaystyle {\widetilde {a}}_{0}={\frac {1}{2L}}\int _{0}^{2L}f({\widetilde {x}})d{\widetilde {x}}$ (6.1.7)

Where the period is represented by 0 to 2L. We are also given that:

$\displaystyle {\frac {\pi }{L}}={\frac {2\pi }{p}}$ (6.1.8)

Rearranging (6.1.8) allows us to solve for L:

$\displaystyle L={\frac {p}{2}}$ (6.1.9)

Multiplying (6.1.9) by 2 allows us to find the period of $\displaystyle a_{0}$, as follows:

$\displaystyle 2L=p$     (6.1.10)

This shows that $\displaystyle a_{0}$ is indeed a periodic function with a period of $\displaystyle p$. This also shows that at any given $\displaystyle x$ value or period throughout the periodic function, $\displaystyle a_{0}$ holds its constant value.